
The radius of the Earth is 6370 km and the radius of mars is 3440 km. What is the acceleration due to Gravity on Mars if the mass of Mars is $1.1$ times the mass of Earth?
Answer
233.1k+ views
Hint: In this solution, we will use the formula of gravitational acceleration on a planet. We will find the gravitational acceleration on the surface of Earth and Mass and take their ratio to determine the acceleration due to gravity on the surface of Mars.
Formula used: In this solution, we will use the following formulae:
Gravitational acceleration on any planet: $g = \dfrac{{GM}}{{{R^2}}}$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$is the radius of the planet.
Complete step by step answer:
We’ve been given that the radius of the Earth is 6370 km and the radius of mars is 3440 km and the mass of Mars is $1.1$ times the mass of Earth.
Let us denote the mass and radius of Earth as ${M_e}\,{\text{and}}\,{{\text{R}}_e}$ and the mass and radius of Mars as ${M_m}\,{\text{and}}\,{{\text{R}}_m}$.
Then the gravitational acceleration on the surface of Earth will be
${g_e} = \dfrac{{G{M_e}}}{{{{\left( {6370} \right)}^2}}}$
And the gravitational acceleration on the surface of Mars will be
${g_m} = \dfrac{{G{M_m}}}{{{{\left( {3440} \right)}^2}}}$
Taking the ratio of the two gravitational acceleration, we get
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{{{M_e}}}{{{M_m}}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Now we’ve been given that the mass of Mars is $1.1$ times the mass of Earth so we can write that mathematically as ${M_m} = 1.1{M_e}$. So, the above equation will be transformed as
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{1}{{1.1}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Which gives us
$\dfrac{{{g_e}}}{{{g_m}}} = 0.265$
Now we know that the gravitational acceleration on the surface of Earth is ${g_e} = 9.81\,m/{s^2}$ so we can find the gravitational acceleration on the surface of Mars as
${g_m} = \dfrac{{9.81}}{{0.265}} = 37\,m/{s^2}$
Note: The dimensions of Mars are different in reality than what is actually mentioned in the question. But this question tests the concepts of gravitational acceleration and its dependence on the mass and radius of a planet.
Formula used: In this solution, we will use the following formulae:
Gravitational acceleration on any planet: $g = \dfrac{{GM}}{{{R^2}}}$ where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$is the radius of the planet.
Complete step by step answer:
We’ve been given that the radius of the Earth is 6370 km and the radius of mars is 3440 km and the mass of Mars is $1.1$ times the mass of Earth.
Let us denote the mass and radius of Earth as ${M_e}\,{\text{and}}\,{{\text{R}}_e}$ and the mass and radius of Mars as ${M_m}\,{\text{and}}\,{{\text{R}}_m}$.
Then the gravitational acceleration on the surface of Earth will be
${g_e} = \dfrac{{G{M_e}}}{{{{\left( {6370} \right)}^2}}}$
And the gravitational acceleration on the surface of Mars will be
${g_m} = \dfrac{{G{M_m}}}{{{{\left( {3440} \right)}^2}}}$
Taking the ratio of the two gravitational acceleration, we get
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{{{M_e}}}{{{M_m}}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Now we’ve been given that the mass of Mars is $1.1$ times the mass of Earth so we can write that mathematically as ${M_m} = 1.1{M_e}$. So, the above equation will be transformed as
$\dfrac{{{g_e}}}{{{g_m}}} = \dfrac{1}{{1.1}} \times {\left( {\dfrac{{3440}}{{6370}}} \right)^2}$
Which gives us
$\dfrac{{{g_e}}}{{{g_m}}} = 0.265$
Now we know that the gravitational acceleration on the surface of Earth is ${g_e} = 9.81\,m/{s^2}$ so we can find the gravitational acceleration on the surface of Mars as
${g_m} = \dfrac{{9.81}}{{0.265}} = 37\,m/{s^2}$
Note: The dimensions of Mars are different in reality than what is actually mentioned in the question. But this question tests the concepts of gravitational acceleration and its dependence on the mass and radius of a planet.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

